Word Problem

I'm the mathematics coordinator of 3rd and 4th grades of my school in Chile. We are using the My Pals are Here Series. But I have a doubt, concerned to the fact that even though we are focused on solving word problems using the models methology and it makes sense to the girls when we are working together, they are still having problems when they are working alone, specially during tests. They start right away with operations but most of the time it is wrong, because they didn't visualize the entire problem. Do you have any suggestion? Should we continue with next chapter or work more on problem solving with additions and subtractions?

Question posted by Paula, a mathematics co-ordinator in Chile

As you have mentioned, when the students start straight away with the operations they are often wrong. They need to comprehend the problems well. Drawing a model will help them understand how the information are related. In simple one-step problem, it may not necessary to do so. But in a problem with a lot of information, this becomes essential for average students. Otherwise although they can read the word, they do not comprehend the information.

Also in multi-step problems, the students may not have the ability to monitor their thinking. This is metacognition. When we teach word problems, we should model and coach rather than explain. That way, we help them in developing the ability to think through the many steps in a problem.

Speed Problem...Again

A car needs 7 hours to travel from Town X to Town Y. A motorcycle needs 8 hours to travel from Town Y to Town X. The car leaves Town X for Town Y and the motorcycles leaves from Town Y to Town X at the same time. How long will it take for the car and the motorcycle to meet?

Angie

Speed Problems are frequently brought up. There are earlier entries discussing Speed Problems. See below.

So, how long will it take for the car and the motorcyle to meet. The standard joke is that we hope they don't!

That aside, we need to assume that the speed of the two vehicles are constant. If that is so then in an hour, the car travels 1/7 the distance in an hour and the motorcycle travels 1/8 the distance in an hour. The problem is solved when the distance travelled by the car and motorcyle add up to 1 whole. In an hour, total distance covered by both is (1/7 + 1/8) of XY. This works out to 15/56 of XY. In 2 hours, it is (2/7 + 2/8) of XY or 30/56 of XY. In 3 hours, 45/56. In 4 hours, 60/56. They would have passed each other in 4 hours. Can I leave it to you to complete the last step of the solution. It is by no means trivial but there are enough leads already.

Request for Presentation Slides

I am a 5th grade teacher in Fayetteville, NC. I had the amazing opportunity to attend your session at the NCTM Conference in Washington, DC a few weeks ago and was truly inspired! If possible, would you be able to email me a copy of your handouts and powerpoints used in your session? I would greatly appreciate it! Thank you for your time and amazing inspiration!

Laura, an American teacher

The presentation slides at the NCTM Annual Meeting & Exposition are available at http://math.nie.edu.sg/t3/downloads-conference.htm Look for the conference that you are interested in and click on the pdf. The slides should download. The slides for my other presentations are also available here.

A Problem from a Hong Kong School

A mom shared that her P1 daughter in Hong Kong was posed this mathematics problem:

AB + B = BA

A = ________B = ________

How do you teach a P1 child algebra?

Lily

In formal algebra, ab means the product of a and b. I do not think that is what the problem is about. In this problem, I believe A and B represent digits and AB is a two-digit number which when added to B gives a two-digit number which has the tens and ones digit of the original number (AB) reversed. In this case the letters are not used in the same convention as formal algebra. I would recommend that this problem is presented orally to grade one children. The teacher might say, "Each letter (or shape) stands for a digit. The same letter stands for the same digit. Different letters stand for different digits. I have a number, the tens digit is A and the ones digit is B (Teacher writes down AB). When it is added to a number B (Teacher writes down AB + B), the total is a number with B as its tens digit and A as its ones digit. (Teacher writes down AB + B = BA) Find the digits A and B."

I would illusrate with an example say 12. What is A in this illustration? What is B? A is 1 and B is 2. So, in this problem AB + B (which is 12 + 2) is supposed to be BA (21). But it is not, right? So AB is not 12. I suppose guess and check is the best strategy for grade one children to use.

Advanced students or older students may reason this way: Is A an odd or even digit? Yes, it must be an even digit. Why? Did you notice that B + B = A. Sure it could be 1A as well. But not 2A or 3A or 4A and so on, right? Since the final sum is BA, AB + B must involve renaming. Why? Otherwise the tens digit in the sum is the A, isn't it? So B must be 6, 7, 8 or 9. And A is one less than B. Think about this one! Hence, A is 5 when B is 6, A is 7 when B is 8 - both not possible. Why?

Hence, A = 6 and B = 7 or A = 8 and B = 9. Checking 67 + 7 = 74 and 89 + 9 = 98. I think the solution is A = 8 and B = 9.

Secondary students may solve it algebraically: 10A + B + B = 10B + A or 9A = 8B or the ratio of A : B = 8 : 9. For digits, A has to be 8 and B has to be 9.

I don't think P1 children are expected to do the algebraic solution or even the reasoning based on number properties. They are most likely able to solve it by guess-and-check.

Another Speed Problem

I have a word problem about speed that I cannot explain to my pupils easily.

Car A and Car B left Town X for Town Y at the same time. Car A was travelling at an average speed of 80 km/h and Car B was travelling at an average speed of 60 km/h. Car A was leading Car B by 8 km for every 1/6 of the distance from Town X to Town Y. Find the distance between Town X and Town Y. One of my friends solve it like this: 6 units x 8 km = 48 km and 80 km - 60 km = 20 km. 80 : 20 = 4. Hence, the distance from Town X to Town Y = 4 x 48 = 192 km. However, the solution is too difficult for my pupils.

Charmaine's Suggestion:

Car A travels 20km more than Car B in an hour. (80-60)

Since Car A leads Car B by 8km for 1/6 of the journey, it leads by a total of 8*6 =48km for the entire journey (6/6).

Thus, time taken for the whole journey by A: 48/20hours = 12/5 hours (leave in simplest improper for easier calculation later...)

Distance between X and Y is thus 12/5*80 = 192km.

Seow's Suggestion

http://farm4.static.flickr.com/3648/3457754184_35d7fcdc47_o.jpg

Speed Problem

I teach maths in Grade 6. There is a word problem in the maths book that I cannot teach using the model method that is usually more suitable for my pupils. The problem: A car and a lorry were travelling towards Town A. The car overtook the lorry when they were 90 km away from Town A. The car arrived at Town A 1/2 h earlier than the lorry while the lorry was 30 km away. Find the average speed of the lorry. Find the average speed of the car.

Gita

Please do not think that the model method can be used for all problems. That is not the idea. We want students to learn a variety of heuristics and they should apply it accordingly. Speed problems are rarely solved using the model method. A line diagram is more useful. Please see any Singapore Grade 6 books for such line diagrams.

In the problem posed, the lorry took 1/2 h to finish the last 30 km. So the average speed of the lorry is easily found (60 km/h). With 90 km to go, both the car and the lorry has travelled the same distance from the spot where the lorry started - that was when the car overtook the lorry. The car must be faster but started later.

I want to suspend the solution for a while. I invite readers to continue to solve the problem. Post further question if necessary.

I have a word problem: The price of a pen was $5, The price of a pencil was $2. Miss Lee bought a number of pens and pencils for $26. How many pens and pencils did she buy?

The day before yesterday I taught my pupils this: I made a list for the pens : 1 x $5 = $5 , 2 x $5 = $10 and so on. And also a list for the pencils : 1 x $2 = $2 and so on. Miss Lee bought 4 pens and 3 pencils because (4 x $5) + (3 x $2) = $26. However, I realize that this method cannot be used in big numbers.

Merry

The method you used is called make a list. It is a common problem-solving heuristic. Please continue to use it with your younger students. I wonder if Miss Lee could also buy 2 pens and some pencils. I know 5 pens is not possible because the money left is an odd number $1 and the price of a pencil is $2. Similarly, 1 pen or 3 pens are not possible. Students learn reasoning.

If students know algebra, they can set up equation 5x + 2y =26 where x is the number of pens and y is the number of pencils. As there is only one condition, you still need to use guess-and-check to solve this equation. Unless the problem says something about x + y.

If the value is not 26 but larger then the equation is 5x + 2y = k where k is the large number. A graph can be plotted and possible solutions seen on the graph. (With larger k the number of solutions increases).

For the young students, the method you use is probably the best. When they are older, they will learn to solve the same problem with larger k values.